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Actual vapor power cycles

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Actual vapor power cycles

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    2. Actual vapor power cycles Actual cycle depart from ideal cycle because of: Frictional irreversibilities (MAJOR) and heat loss from the turbine Frictional irreversibilities (MAJOR) and heat loss from the pump Pressure drops due to friction and heat losses in the piping, esp.: Boiler exit to turbine inlet Pump exit to boiler inlet Frictional pressure drop and subcooling (to avoid cavitation in the pump) in the condenser.

    3. Actual vapor power cycles

    4. Work done by and exit state of an actual turbine Example 7.4Example 7.4

    5. Work done by and exit state of an actual pump Example 7.4Example 7.4

    6. Analysis of an actual Rankine cycle with irreversibilities in turbine and pump Water is the working fluid in a Rankine cycle. Superheated vapour enters the turbine which has an isentropic efficiency of 85% at 8 MPa, 480oC. The condenser pressure is 8 kPa. The pump has an isentropic efficiency of 70%. The net power output of the cycle is 100 MW. Calculate the quality of steam at turbine exit efficiency of the cycle mass flow rate of steam

    7. EES solution procedure Power=100e3; Pboil=8000; Pcond=8; isenturb=0.85; isenpump=0.75; h1=enthalpy('steam', P=Pboil,T=480); "Table A-6" s1=entropy('steam',P=Pboil,T=480); "Table A-6" x2s=quality('steam', P=Pcond,s=s1); "x2s=(s1-sf)/sfg at P=Pcond from Table A-5: needed to find h2s: this is not the quality at the exit of actual turbine" h3=enthalpy('steam',P=Pcond,x=0); " Table A-5: To find h2s but it is also enthalpy at state 3" hg2=enthalpy('steam',P=Pcond,x=1); "Table A-5: to find h2s" h2s=h3+x2s*(hg2-h3); "h2s=enthalpy('steam',P=Pcond,s=s1);" "Using EES, you could have directly found h2s; last four steps not necessary“ wturb=(h1-h2s)*isenturb; h2=h1-wturb "enthalpy at exit to the turbine" x2=quality('steam',h=h2,P=Pcond); " x2=(h2-hf)/hfg at P=Pcond from Table A-5: Ans b” v3=volume('steam',P=Pcond,x=0); "Table A-5: needed to calculate pump work" wpump=v3*(Pboil-Pcond)/isenpump; h4=h3+wpump; eta=(wturb-wpump)/(h1-h4);"Ans a" mdot=Power/(wturb-wpump);"Ans c"

    8. EES solution With irreversibilities in turbine and pump eta=0.3365 h1=3349 [kJ/kg] h2=2273 h2s=2083 h3=173.7 [kJ/kg] h4=185.3 hg2=2576 [kJ/kg] isenpump=0.7 isenturb=0.85 mdot=93.93 Pboil=8000 Pcond=8 Power=100000 s1=6.659 [kJ/kg-K] v3=0.001008 [m^3/kg] wpump=11.51 wturb=1076 x2=0.8737 x2s=0.7946

    9. Effect of superheat

    10. Ideal Rankine cycle with and without superheat

    11. Effect of superheat through an example Steam enters a turbine of a simple vapor power plant with a pressure of 10 MPa and temperature T, and expands adiabatically to 6 kPa. The isentropic turbine efficiency is 85%. Saturated liquid exits the condenser at 6 kPa and the isentropic pump efficiency is 82%. Plot the the turbine exit quality and cycle efficiency for T=312-700oC.

    12. EES solution procedure using SOLVE TABLE P1=10000; P2=6; isenturb=0.85; isenpump=0.82; h1=enthalpy('steam', P=P1,T=T); s1=entropy('steam',P=P1,T=T); h2s=enthalpy('steam',P=P2,s=s1); "Using EES, you can find h2s without using x2s“ h2=h1-wturb x2=quality('steam',h=h2,P=P2); "Ans b" v3=volume('steam',P=P2,x=0); wturb=(h1-h2s)*isenturb; wpump=v3*(P1-P2)/isenpump; h3=enthalpy('steam',P=P2,x=0); h4=h3+wpump; eta=(wturb-wpump)/(h1-h4);"Ans a"

    13. Efficiency and superheat

    14. Exit quality and superheat

    15. Superheat for higher exit quality: another example Water is the working fluid in an ideal Rankine cycle. Superheated vapour enters the turbine at 8 MPa and at a temperature such that the quality at the exit of the turbine is 0.85. The condenser pressure is 8 kPa. Calculate the Temperature of steam at the turbine inlet efficiency of the cycle mass flow rate of steam

    16. EES solution procedure Pboil=8000; Pcond=8; wf$='steam'; xmin=0.85; h2=enthalpy(wf$, P=Pcond,x=xmin); "Table A 5" s2=entropy(wf$, P=Pcond,x=xmin); "Table A 5" h1=enthalpy(wf$, P=Pboil,s=s2); "Table A-6" T1=temperature(wf$, P=Pboil,s=s2); "Table A-6: Ans a" v3=volume(wf$,P=Pcond,x=0); "Table A-5" h3=enthalpy(wf$,P=Pcond,x=0); "Table A-5" wturb=h1-h2; "Table A-5" wpump=v3*(Pboil-Pcond); "Using expression for reversible steady flow work" h4=h3+wpump; eta=(wturb-wpump)/(h1-h4); “Ans b”

    17. EES SOLUTION eta=0.419 h1=3696 [kJ/kg] h2=2216 [kJ/kg] h3=173.7 [kJ/kg] h4=181.8 [kJ/kg] Pboil=8000 [kPa] Pcond=8 [kPa] s2=7.082 [kJ/kg-K] T1=622.8 [C] v3=0.001008 [m^3/kg] wpump=8.06 [kJ] wturb=1481 [kJ] xmin=0.85

    18. Superheating: summary Superheating improves both efficiency and the turbine exit quality (T?? ??, x?). However, the maximum temperature of superheat is limited by metallurgical considerations (current state of the art allows about 700 OC).

    19. Reheat

    20. Effect of increasing boiler pressure If the maximum temperature of superheat is fixed from metallurgical considerations, on increasing the boiler pressure (see boardwork): Efficiency increases?high boiler pressures desirable BUT, turbine exit quality decreases?Ideal Rankine cycles cannot be used at high boiler pressures.

    21. Reheat: a solution that allows for high boiler pressures at high turbine exit quality

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